Geometric Sig.$\frac{d}{dt}(|\lVert \boldsymbol{r}(t)\rVert )=\frac{\boldsymbol{r}(t).\frac{d}{dt}(\boldsymbol{r}(t))}{\lVert\boldsymbol{r}(t)\rVert}$ Suppose $\boldsymbol{r}\not=0$, what's the geometric derivation/significance of the above, the question is a show that, I could (probably) just use definitions.
Instead I thought about when it'd be zero, if the derivative of the magnitude of a vector is 0, it's never changing length, it could be doing nothing (so r`(t)=0) or it could be moving in a circle/sphere/whatever is above that around the origin, thus never changing length.
If the vector starts at the origin and is moving away from the origin then in the same direction as it's position (vector) .... I can see there's something nice here, not quite sure what.
Anyway I'm faffing with poorly drawn diagrams is there a nice way of doing this, I'd like to show this geometrically, I've "done" the case when change in length is zero. 
Perhaps I am wrong and there isn't something nice here, but the circle/spherical thing is just a special case, then I shall defer to from definitions.
If there is something nice, what is it?
Thanks
I've not used Latex much in the text because I'd rather use pictures and a thought process, if you want me to Latex what 'lil math there is, add a comment.
using $r=\boldsymbol{r}(t)$ and $|r| = \lVert\boldsymbol{r}(t)\rVert$ and r` as the derivative of r
$$\frac{d}{dt}(|r|) = \frac{r.r'}{|r|}$$
implies (iff? why not, why implies? Not entirely confident here)
$$|r|\frac{d}{dt}(|r|)=r.r'=|r||r'|cos(\theta)$$
where theta is the angle between r and r`
Now 
$$\frac{d}{dt}(|r|)=|r'|cos(\theta)$$
"The rate of change of length from the origin to a point on the curve is the amount of it's magnitude of it's rate of change in the direction from the origin to that point"
Shall I pose this as my own answer?
 A: The vector $d\mathbf r/dt$ can be said to have a component that is radial--along the direction of $\hat{\mathbf r}$--and a component that is angular, or perpendiular to $\hat{\mathbf r}$.
The parallel component is clearly $\hat{\mathbf r}(\hat{\mathbf r} \cdot d\mathbf r/dt)$--this is a basic projection operation.  The magnitude of this component is exactly $d|\mathbf r|/dt$--you should be able to picture this geometrically that any angular component of $d\mathbf r/dt$ cannot change the magnitude of $\mathbf r$ because the angular components are all orthogonal to $\mathbf r$.
A: This is seemingly simple enough. For two time dependent vectors $ a, b$ you have the relation $$ \frac{d}{dt} (a.b) = a.\frac{db}{dt}+ b.\frac{da}{dt}$$ Using this and chain rule you can calculate $ \frac{d}{dt}(r(t).r(t) = 2r(t).\frac{d}{dt}r(t) $ and hence
$$ \frac{d}{dt}\|r(t)\| = \frac{d}{dt}\sqrt{r(t).r(t)} = \frac{1}{2\sqrt{r(t).r(t)}}\frac{d}{dt} (r(t).r(t)) = \frac{r(t)}{\|r(t)\|}.\frac{d}{dt}r(t) $$ This happens because $ \frac{d}{dt}\|r(t) \| $ is the speed, and this relation shows that it is just the magnitude of the velocity, that is the value of the component of velocity in its direction.( $\frac{r(t)}{\|r(t)\|}$ being the unit vector in that direction).
